Commutativity of Lorentz Boosts & Rotations

[parsikoo]

As per group property, one could make a product of gr members e.g. Lorentz boost (B) and rotation R, as they Commutativity is not valid, R.B or B.R, what should be considered and which order should be preferred? Generally it is known R.B1= B2.R .

 

[haushofer]

Nothing is preferred, it depends on what you want to consider. Your question is a bit vague.

 

[parsikoo]

Nothing is preferred, it depends on what you want to consider. Your question is a bit vague.

Let's be more clear, if a particle moves in a single direction in one coordinate e.g. boost, then an observer has a finite rotation, then the transformation in his frame is simply the product of a boost and rotation, then what are the considerations as regards the order?

 

[haushofer]

I still don't get it. If you first perform a boost, and then a rotation, then the order is boost-rotation. Are you perhaps puzzled by active vs passive (i.e. boost the particle or boost the observer)?

 

[parsikoo]

I meant, boost in a frame, brought to a new frame which makes a finite rotation, hope it is clear.

 

[haushofer]

A boost is not a spatial rotation. Maybe it is even easier to visualize it for the Galilei group. A Galilei boost is a spatial translation linear in time. A rotation is a different kind of transformation. In your example it would be boost, and then rotation. You can also first rotate, and then boost. But that will give a different answer, which you can simply see by performing the infinit. transformations on the coordinates.

 

[parsikoo]

Mathematically if you would use iterate many infinitesimal transformations, then the transf. by exp. form would be exp(e1.B+ e2.R) (e=infinitesimal amount), but it would be interesting to see the proof that the outcome of changing the order of R and B would lead to equal transformation

 

[Muphrid]

The order is going to be specified. No one is going to say "boost this way and rotate that way" without explicitly saying which comes first, just for that reason. If they do, they don't have an understanding of the math.

That said, as long as the boost plane and the rotation plane share no common direction, the boost and the rotation commute. But that's rather trivial and obvious. Of course a tx-plane boost won't affect a yz-plane rotation, and vice versa.

 

[parsikoo]

Can we please focus on our specific case.

 

[haushofer]

Mathematically if you would use iterate many infinitesimal transformations, then the transf. by exp. form would be exp(e1.B+ e2.R) (e=infinitesimal amount), but it would be interesting to see the proof that the outcome of changing the order of R and B would lead to equal transformation

You can use the baker- campbell- hausdorff relation for that, and the explicit algebra, to see if that's true.

The algebra ( and also the application of the infinit. transfo. on the coordinates) tells you that rotations and boosts don't commute; the boost parameter is a vector, transforming as a vector under the adjoint action of a rotation. Schematically,

[rot, boost] = boost

 

[TrickyDicky]

parsikoo, maybe you are referring to the fact that proper Lorentz transformations in 3D also contains a rotation of the three axes, because the composition of two boosts is not just a boost but is a boost followed by a rotation (look up also Thomas precesion).